Projections in Banach spaces Dear All, 
I am absolutely lost in the following problem:
Let $P_s, \: s \in [0,1],$ be a uniformly bounded family of projections (idempotents) in a Banach space $X$ such that $P_s P_t = P_{{\rm min}(s,t)}$. Let $Q$ be a bounded linear operator on $X$ such that $QP_s = P_sQ$ for every $s \in [0,1]$ and the function $$s \mapsto P_s Q $$ is continuous in operator norm. Does it follow that the function must be constant (i.e. $P_sQ \equiv P_0Q$)? 
For simplicity, one can also assume that the function $s \mapsto P_s$ is strongly continuous. I can give an affirmative answer only in some trivial situations (finite dimensional case, Hilbert spaces with family of orthogonal projections) but nothing more. 
 A: Here a simple example :
Let X be the cartesian product of $L^{\infty}$ and $L^{1}$ on the
interval $[0,1]$, let $P_{t}$ the canonical projection on the subspace
of functions with support $[0,t]$ and choose $Q(f_{1},f_{2}) =
(0,f_{1})$ .
A: I'm not sure if I like the ultrafilters, so I decided to find some elementary construction. I do not have much imagination for chains of commuting projections either, so let us consider the space of all functions $f:[0,1]\to X$ where $X$ is the space of all sequences and let $P_s$ just keep the values to the left of $s$ and kill the values to the right of $s$. So far so good. The operator $Q$ will just act in each layer separately, so it is going to be an operator in $X$ really. The devil is in the choice of the norm. We need to take an advantage of small support somehow, which calls for considering a sequence of $L^{p_k}$-norms on $[0,1]$ with decreasing $p_k>1$. Then we'll have the full strength of Holder backing us. However, there needs to be some penalty for using norms with small $p_k$, so it is natural to pair each $p_k$ with some norm $N_k$ in $X$, which are increasing fast so that the final norm, which is just the infimum of $\sum_k\|N_k(f_k(t))\|_{L^{p_k}}$ over all decompositions $f=\sum_k f_k$ will have to be a hard trade-off rather than a trivial collapse to a single term. Now the first thing we want is 
$$
N_{k+1}(Qx)\le N_k(x)
$$
This will ensure that for the first small $k$ we will fire with Holder to gain on the power of the length of the interval but for the large $k$, when Holder finally betrays us and the reduction of power becomes useless, we will use
$$
N_k(Qx)\le 2^{-k} N_k(x)
$$
so each faraway term will just take care of itself.
The possibilities for the choice of such family of norms and $Q$ are unlimited but, being (sort of) an analyst, I like the backward shift and weighted spaces, so put
$$
(Qx)_{j}=2^{-j}x_{j+1},
$$
and
$$
N_k(x)=\sum_j 2^{kj}|x_j|.
$$ 
A: I guess that the answer is no in general. More precisely what I consider as the discrete version of your question has a negative answer. I guess that one should be able to find a couterexample to your question by an ultraproduct argument, but I did not check the details.
By discrete version of your question I mean: if $(P_n)_{n \in \mathbb N}$ is a family of uniformly bounded projections on $X$ such that $P_0=0$ and $P_n P_m = P_{\min(n,m)}$, and $Q$ is a (not necessaily bounded) linear map on $X$ that commutes with the $P_n$'s and is such that $\sup_n \|P_n Q -P_{n-1} Q\|<\infty$. Does it follow that $\sup_n \|P_n Q\|<\infty$?
If $(e_n)$ is a Schauder basis in $X$, and take $P_n$ the map $x = \sum_k x_k e_k \mapsto P_n(x) = \sum_{k \leq n} x_k e_k$. Then $P_n$ satisfies your assumption. For a sequence $z_n \in \{{-1,1\}}$, define $Q x_n = z_n x_n$ so that $\|P_n Q - P_{n-1} Q\| = \|P_n - P_{n-1}\|$. Then $\sup_n \|P_n Q\| <\infty$ for every such $Q$ is equivalent to the basis $(e_n)$ being unconditional. It is well known that there exist bases that are not unconditional. This answers negatively the discrete question.
Here is how I would try to deduce a continuous example from a discrete example $X,P_n,Q$~: for any $k$, define  $P_s^{(k)} = P_{[2^k s]}$ and $Q_k = P_{2^k}Q/\|Q P_{2^k}\|$. Then consider a non principal ultrafilter on $\mathbb N$, and construct the following operators on $X^{\mathcal U}$, and the operators $P_s = (P_s^{(k)})_{k \in \mathbb N}$ and $Q_{\mathcal U} = (Q_k)_{k \in \mathbb N}$. Then $\|Q_{\mathcal U}\|=1$, so that $P_0 Q_{\mathcal U} = 0 \neq Q_{\mathcal U}= P_1 Q_{\mathcal U}$. One probably needs to assume something more on $Q$ to prove that $s \mapsto P_s Q$ is continuous. I leave this to you.
